The common roots of the equations z 3+ 2z 2+ 2z + 1 = 0 and z 2017 + z 2018 + 1 = 0 are
ω, ω 2
Finding Common Roots of Polynomial Equations
We are asked to find the common roots of two polynomial equations:
\( z^3 + 2z^2 + 2z + 1 = 0 \)
\( z^{2017} + z^{2018} + 1 = 0 \)
To find the common roots, we first need to find the roots of the first equation and then check which of these roots satisfy the second equation.
Solving the First Equation: \( z^3 + 2z^2 + 2z + 1 = 0 \)
This is a cubic equation. We can try testing simple values like \( z = -1 \) or \( z = 1 \) to see if they are roots.
Let's test \( z = -1 \):
\[ (-1)^3 + 2(-1)^2 + 2(-1) + 1 = -1 + 2(1) - 2 + 1 = -1 + 2 - 2 + 1 = 0 \]Since the expression evaluates to 0, \( z = -1 \) is a root of the first equation. This means \( (z+1) \) is a factor of the polynomial \( z^3 + 2z^2 + 2z + 1 \).
We can perform polynomial division to find the other factor:
\( (z^3 + 2z^2 + 2z + 1) \div (z+1) \)
Using synthetic division or long division, we find:
\[ z^3 + 2z^2 + 2z + 1 = (z+1)(z^2 + z + 1) \]So, the first equation can be written as:
\[ (z+1)(z^2 + z + 1) = 0 \]The roots are found by setting each factor to zero:
\( z+1 = 0 \implies z = -1 \)
\( z^2 + z + 1 = 0 \)
The quadratic equation \( z^2 + z + 1 = 0 \) is related to the complex cube roots of unity. The roots of this equation are given by the quadratic formula:
\[ z = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]These are the complex cube roots of unity, commonly denoted by \( \omega \) and \( \omega^2 \), where \( \omega = e^{i 2\pi/3} = \frac{-1 + i\sqrt{3}}{2} \) and \( \omega^2 = e^{i 4\pi/3} = \frac{-1 - i\sqrt{3}}{2} \). We know that \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).
Therefore, the roots of the first equation \( z^3 + 2z^2 + 2z + 1 = 0 \) are \( -1 \), \( \omega \), and \( \omega^2 \).
Checking the Roots in the Second Equation: \( z^{2017} + z^{2018} + 1 = 0 \)
Now we test each of the roots \( -1, \omega, \omega^2 \) in the second equation.
Test \( z = -1 \)
Substitute \( z = -1 \) into \( z^{2017} + z^{2018} + 1 \):
\[ (-1)^{2017} + (-1)^{2018} + 1 \]Since 2017 is odd, \( (-1)^{2017} = -1 \). Since 2018 is even, \( (-1)^{2018} = 1 \).
\[ -1 + 1 + 1 = 1 \]Since the result is 1 (not 0), \( z = -1 \) is not a root of the second equation. Thus, \( -1 \) is not a common root.
Test \( z = \omega \)
Substitute \( z = \omega \) into \( z^{2017} + z^{2018} + 1 \). We use the property \( \omega^3 = 1 \).
For \( z^{2017} \): We divide 2017 by 3.
\[ 2017 = 3 \times 672 + 1 \]So, \( \omega^{2017} = \omega^{3 \times 672 + 1} = (\omega^3)^{672} \cdot \omega^1 = 1^{672} \cdot \omega = 1 \cdot \omega = \omega \).
For \( z^{2018} \): We divide 2018 by 3.
\[ 2018 = 3 \times 672 + 2 \]So, \( \omega^{2018} = \omega^{3 \times 672 + 2} = (\omega^3)^{672} \cdot \omega^2 = 1^{672} \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2 \).
Substitute these back into the second equation:
\[ \omega^{2017} + \omega^{2018} + 1 = \omega + \omega^2 + 1 \]From the properties of cube roots of unity, we know that \( 1 + \omega + \omega^2 = 0 \).
So, \( \omega + \omega^2 + 1 = 0 \). This means \( z = \omega \) is a root of the second equation. Thus, \( \omega \) is a common root.
Test \( z = \omega^2 \)
Substitute \( z = \omega^2 \) into \( z^{2017} + z^{2018} + 1 \). We use the property \( (\omega^2)^3 = \omega^6 = (\omega^3)^2 = 1^2 = 1 \).
For \( z^{2017} \): \( (\omega^2)^{2017} = \omega^{2 \times 2017} = \omega^{4034} \).
We divide 4034 by 3.
\[ 4034 = 3 \times 1344 + 2 \]So, \( \omega^{4034} = \omega^{3 \times 1344 + 2} = (\omega^3)^{1344} \cdot \omega^2 = 1^{1344} \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2 \).
For \( z^{2018} \): \( (\omega^2)^{2018} = \omega^{2 \times 2018} = \omega^{4036} \).
We divide 4036 by 3.
\[ 4036 = 3 \times 1345 + 1 \]So, \( \omega^{4036} = \omega^{3 \times 1345 + 1} = (\omega^3)^{1345} \cdot \omega^1 = 1^{1345} \cdot \omega = 1 \cdot \omega = \omega \).
Substitute these back into the second equation:
\[ (\omega^2)^{2017} + (\omega^2)^{2018} + 1 = \omega^2 + \omega + 1 \]Again, we know that \( 1 + \omega + \omega^2 = 0 \).
So, \( \omega^2 + \omega + 1 = 0 \). This means \( z = \omega^2 \) is a root of the second equation. Thus, \( \omega^2 \) is a common root.
Summary of Common Roots
The roots of the first equation are \( -1, \omega, \omega^2 \). The roots of the second equation that are also roots of the first equation are \( \omega \) and \( \omega^2 \).
Therefore, the common roots of the two equations are \( \omega \) and \( \omega^2 \).
| Root Candidate | Is root of \( z^3 + 2z^2 + 2z + 1 = 0 \) ? | Is root of \( z^{2017} + z^{2018} + 1 = 0 \) ? | Is a Common Root? |
|---|---|---|---|
| \( -1 \) | Yes | No (Result is 1) | No |
| \( \omega \) | Yes | Yes (Result is \( \omega + \omega^2 + 1 = 0 \)) | Yes |
| \( \omega^2 \) | Yes | Yes (Result is \( \omega^2 + \omega + 1 = 0 \)) | Yes |
Revision Table: Complex Roots of Unity
| Term | Definition/Properties |
|---|---|
| Cube roots of unity | Solutions to \( z^3 = 1 \). These are \( 1, \omega, \omega^2 \). |
| \( \omega \) | A complex cube root of unity, \( e^{i 2\pi/3} = \frac{-1 + i\sqrt{3}}{2} \). |
| \( \omega^2 \) | The other complex cube root of unity, \( e^{i 4\pi/3} = \frac{-1 - i\sqrt{3}}{2} \). Note \( \omega^2 = \bar{\omega} \). |
| Key Property 1 | \( \omega^3 = 1 \) (and \( (\omega^k)^3 = 1 \) if \( k \) is an integer multiple of 3, in general \( \omega^{3n} = 1 \)) |
| Key Property 2 | \( 1 + \omega + \omega^2 = 0 \) |
| Powers of \( \omega \) | \( \omega^n = \omega^{n \pmod 3} \). For example, \( \omega^4 = \omega^1 = \omega \), \( \omega^5 = \omega^2 \), \( \omega^6 = \omega^0 = 1 \). |
| Powers of \( \omega^2 \) | \( (\omega^2)^n = \omega^{2n} \). Then use \( \omega^{2n} = \omega^{2n \pmod 3} \). For example, \( (\omega^2)^2 = \omega^4 = \omega \), \( (\omega^2)^3 = \omega^6 = 1 \), \( (\omega^2)^4 = \omega^8 = \omega^2 \). |
Additional Information on Polynomial Roots
Finding the roots of a polynomial equation is a fundamental concept in algebra. For cubic equations, techniques include factoring, using the rational root theorem to find potential integer or rational roots, and then dividing the polynomial to find a quadratic factor. The roots of the quadratic factor can be found using the quadratic formula.
Complex numbers, like the complex cube roots of unity \( \omega \) and \( \omega^2 \), often appear as roots of polynomial equations with real coefficients. If a polynomial with real coefficients has a complex root \( a + bi \) where \( b \ne 0 \), then its conjugate \( a - bi \) must also be a root.
In this problem, recognizing the roots of \( z^2+z+1=0 \) as \( \omega \) and \( \omega^2 \) is key. The properties \( \omega^3=1 \) and \( 1+\omega+\omega^2=0 \) are essential for simplifying expressions involving high powers of \( \omega \) and \( \omega^2 \).
When looking for common roots of two equations, you find the roots of one equation and substitute them into the other equation. If a root satisfies both equations, it is a common root.
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